A curve is defined by the parametric equations $x=t^3+1$ and $y=t^4-4$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{4}{9t}$ (Choice B) B $\dfrac{4}{9t^2}$ (Choice C) C $\dfrac{3t^3+3}{4t}$ (Choice D) D $\dfrac{4t}{3}$
Explanation: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\dfrac{4t}{3}$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(\dfrac{4t}{3}\right)}{\dfrac{d}{dt}(t^3+1)} \\\\ &=\dfrac{\dfrac{4}{3}}{3t^2} \\\\ &=\dfrac{4}{9t^2} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=\dfrac{4}{9t^2}$.